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Forum Geometricorum
Volume 2 (2002) 183–185.
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FORUM GEOM
ISSN 1534-1178
An Application of Thébault’s Theorem
Wilfred Reyes
Abstract. We prove the “Japanese theorem” as a very simple corollary of Thébault’s
theorem.
Theorem 1 below is due to the French geometer Victor Thébault [8]. See Figure
1. It had been a long standing problem, but a number of proofs have appeared
since the early 1980’s. See, for example, [7, 6, 1], and also [5] for a list of proofs
in Dutch published in the 1970’s. A very natural and understandable proof based
on Ptolemy’s theorem can be found in [3].
Theorem 1 (Thébault). Let E be a point on the side of triangle ABC such that
∠AEB = θ. Let O1 (r1 ) be a circle tangent to the circumcircle and to the segments
EA, EB. Let O2 (r2 ) be also tangent to the circumcircle and to EA, EC. If I(ρ)
is the incircle of ABC, then
O1 I
= tan2 2θ ,
(1.1) I lies on the segment O1 O2 and
IO2
(1.2) ρ = r1 cos2 θ2 + r2 sin2 2θ .
A
A
B
O2
Id
E
Ic
Ib
I
Ia
O1
B
C
E
Figure 1
C
D
Figure 2
Theorem 2 below is called the “Japanese Theorem” in [4, p.193]. See Figure
2. A very long proof can be found in [2, pp.125–128]. In this note we deduce the
Japanese Theorem as a very simple corollary of Thébault’s Theorem.
Publication Date: December 30, 2002. Communicating Editor: Paul Yiu.
184
W. Reyes
Theorem 2. Let ABCD be a convex quadrilateral inscribed in a circle. Denote
by Ia (ρa ), Ib (ρb ), Ic (ρc ), Id (ρd ) the incircles of the triangles BCD, CDA, DAB,
and ABC.
(2.1) The incenters form a rectangle.
(2.2) ρa + ρc = ρb + ρd .
Proof. In ABCD we have the following circles: Ocd (rcd ), Oda (rda ), Oab (rab ),
and Obc (rbc ) inscribed respectively in angles AEB, BEC, CED, and DEA, each
tangent internally to the circumcircle. Let ∠AEB = ∠CED = θ and ∠BEC =
∠DEA = π − θ.
A
B
Ocd
Id
Ic
Oda
E
Obc
Ib
Ia
C
D
Oab
Figure 3
Now, by Theorem 1, the centers Ia , Ib , Ic , Id lie on the lines Oda Oab , Oab Obc ,
Obc Ocd , Ocd Oda respectively. Furthermore,
Obc Ic
θ
π−θ
Oda Ia
= cot2 ,
=
= tan2
Ia Oab
Ic Ocd
2
2
Ocd Id
θ
Oab Ib
=
= tan2 .
Ib Obc
Id Oda
2
From these, we have
Obc Ib
Oda Ia
=
,
Ia Oab
Ib Oab
Oda Id
Obc Ic
=
,
Ic Ocd
Id Ocd
Oab Ib
Ocd Ic
=
,
Ib Obc
Ic Obc
Ocd Id
Oab Ia
=
.
Id Oda
Ia Oda
An application of Thébault’s theorem
185
These proportions imply the following parallelism:
Ia Ib //Oda Obc ,
Ib Ic //Oab Ocd ,
Ic Id //Obc Oda ,
Id Ia //Ocd Oab .
As the segments Ocd Oab and Oda Obc are perpendicular because they are along the
bisectors of the angles at E, Ia Ib Ic Id is an inscribed rectangle in Oab Obc Ocd Oda ,
and this proves (2.1).
Also, the following relation results from (1.2):
θ
θ
ρa + ρc = (rab + rcd ) cos2 + (rda + rbc ) sin2 .
2
2
This same expression is readily seen to be equal to ρb + ρd as well. This proves
(2.2).
References
[1] S. Dutta, Thébault’s problem via euclidean geometry, Samayā, 7 (2001) number 2, 2–7.
[2] H. Fukagawa and D. Pedoe, Japanese Temple Geometry, Charles Babbage Research Centre,
Manitoba, Canada, 1989.
[3] S. Gueron, Two applications of the generalized Ptolemy theorem, Amer. Math. Monthly, 109
(2002) 362–370.
[4] R. A. Johnson, Advanced Euclidean Geometry, 1925, Dover reprint.
[5] F. M. van Lamoen, Hyacinthos message 5753, July 2, 2002.
[6] R. Shail, A proof of Thébault’s theorem, Amer. Math. Monthly, 108 (2001) 319–325.
[7] K. B. Taylor, Solution of Problem 3887, Amer. Math. Monthly, 90 (1983) 486–187.
[8] V. Thébault, Problem 3887, Amer. Math. Monthly, 45 (1948) 482–483.
Wilfred Reyes: Departamento de Ciencias Básicas, Universidad del Bı́o-Bı́o, Chillán, Chile
E-mail address: wreyes@ubiobio.cl